This will be an incomplete answer, but in case no one else comes along:
The idea of the map $H(s) = 1 + e^s$ is that you are mapping points $s$ on a circle of radius $R_n = 2n\pi$ to another complex number $\zeta = 1+ e^x e^{iy}$, or say $\zeta - 1 = e^x e^{iy}$. From this form, you can see that the $x$ component of the coordinate $s$, which lies on the circle, becomes the magnitude of this new number $\zeta -1$. The $y$ component, $iy$, becomes the new phase. In a way you're mapping a circle onto another circle, since the components of one circle $C_n$ become the magnitude and angle of another circle-looking curve.
They first ask you to consider the region $0 < x \le x_0$. If $R_n = 2n\pi$ is big, the circle will hardly vary at all from $x=0$ to $x = x_0$, so you can think of the region as a rectangle in the complex plane with top and bottom at $y = i \pm 2n\pi$. Then you can see as you traverse the rectangle from $x = 0$ to $x = x_0$, the number $\zeta - 1$ goes from $e^x e^{iy} = e^{iy} = e^{i2n\pi}$ to $e^{x_0} e^{i2n\pi}$. The key is that the radius of this circle happens to be a multiple of $2\pi$, so then $\zeta - 1$ goes from $1$ to $e^{x_0}$. And $e^{x_0}$ will always be greater than 1, which is where they get the first part $\left|{1 + e^s}\right| \ge 2$.
Then in the region $x < 0$, the number $ \zeta - 1$ will be $e^x e^{iy} = e^{R_n \cos{\phi}} e^{iR_n \sin{\phi}} = e^{2n\pi \cos{\phi}} e^{i 2n\pi \sin{\phi}}$. Then if $ y = 2n\pi \sin{\phi}$ goes from $ 2n\pi $ to $(2n-1)\pi$, the number $\zeta - 1$ will make a half revolution, since its phase $y$ went down by $\pi$. From there, you are essentially at the conclusion.
